System of Particles and Rotational Motion - Result Question 58

62. A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2 , the kinetic energy of the mass will

[2011M]

(a) remain constant

(b) increase by a factor of 2

(c) increase by a factor of 4

(d) decrease by a factor of 2

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Answer:

Correct Answer: 62. (c)

Solution:

(c) K.E. $=\frac{L^{2}}{2 I}$

The angular momentum $L$ remains conserved about the centre.

That is, $L=$ constant.

$I=mr^{2}$

$\therefore$ K.E. $=\frac{L^{2}}{2 mr^{2}}$

In 2 nd case, K.E. $=\frac{L^{2}}{2(mr^{\prime 2})}$

But $r^{\prime}=\frac{r}{2}$

$\therefore K . E^{\prime}=\frac{L^{2}}{2 m \cdot \frac{r^{2}}{4}}=\frac{4 L^{2}}{2 mr^{2}} \Rightarrow$ K.E.’ $=4$ K.E.

$\therefore$ K.E. is increased by a factor of 4 .

When no torque is applied angular momentum remains constant.

$I_1 \omega_1=I_2 \omega_2 \Rightarrow \frac{\omega_2}{\omega_1}=\frac{I_1}{I_2}$

$\therefore \frac{E_2}{E_1}=\frac{\frac{1}{2} I_2 \omega_2^{2}}{\frac{1}{2} I_1 \omega_1^{2}}=\frac{I_2}{I_1}(\frac{I_1}{I_2})^{2}=\frac{I_2}{I_2}$